3.816 \(\int \frac {x^5}{\sqrt {a+b x^4}} \, dx\)

Optimal. Leaf size=53 \[ \frac {x^2 \sqrt {a+b x^4}}{4 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}} \]

[Out]

-1/4*a*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(3/2)+1/4*x^2*(b*x^4+a)^(1/2)/b

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Rubi [A]  time = 0.03, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {275, 321, 217, 206} \[ \frac {x^2 \sqrt {a+b x^4}}{4 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/Sqrt[a + b*x^4],x]

[Out]

(x^2*Sqrt[a + b*x^4])/(4*b) - (a*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*b^(3/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {x^5}{\sqrt {a+b x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,x^2\right )\\ &=\frac {x^2 \sqrt {a+b x^4}}{4 b}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{4 b}\\ &=\frac {x^2 \sqrt {a+b x^4}}{4 b}-\frac {a \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )}{4 b}\\ &=\frac {x^2 \sqrt {a+b x^4}}{4 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 53, normalized size = 1.00 \[ \frac {x^2 \sqrt {a+b x^4}}{4 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/Sqrt[a + b*x^4],x]

[Out]

(x^2*Sqrt[a + b*x^4])/(4*b) - (a*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*b^(3/2))

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fricas [A]  time = 0.78, size = 101, normalized size = 1.91 \[ \left [\frac {2 \, \sqrt {b x^{4} + a} b x^{2} + a \sqrt {b} \log \left (-2 \, b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right )}{8 \, b^{2}}, \frac {\sqrt {b x^{4} + a} b x^{2} + a \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x^{2}}{\sqrt {b x^{4} + a}}\right )}{4 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(2*sqrt(b*x^4 + a)*b*x^2 + a*sqrt(b)*log(-2*b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a))/b^2, 1/4*(sqrt(b*
x^4 + a)*b*x^2 + a*sqrt(-b)*arctan(sqrt(-b)*x^2/sqrt(b*x^4 + a)))/b^2]

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giac [A]  time = 0.19, size = 44, normalized size = 0.83 \[ \frac {\sqrt {b x^{4} + a} x^{2}}{4 \, b} + \frac {a \log \left ({\left | -\sqrt {b} x^{2} + \sqrt {b x^{4} + a} \right |}\right )}{4 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^4+a)^(1/2),x, algorithm="giac")

[Out]

1/4*sqrt(b*x^4 + a)*x^2/b + 1/4*a*log(abs(-sqrt(b)*x^2 + sqrt(b*x^4 + a)))/b^(3/2)

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maple [A]  time = 0.01, size = 43, normalized size = 0.81 \[ \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{4 b}-\frac {a \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^4+a)^(1/2),x)

[Out]

1/4*x^2*(b*x^4+a)^(1/2)/b-1/4*a/b^(3/2)*ln(x^2*b^(1/2)+(b*x^4+a)^(1/2))

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maxima [A]  time = 2.90, size = 81, normalized size = 1.53 \[ \frac {a \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x^{4} + a}}{x^{2}}}{\sqrt {b} + \frac {\sqrt {b x^{4} + a}}{x^{2}}}\right )}{8 \, b^{\frac {3}{2}}} - \frac {\sqrt {b x^{4} + a} a}{4 \, {\left (b^{2} - \frac {{\left (b x^{4} + a\right )} b}{x^{4}}\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

1/8*a*log(-(sqrt(b) - sqrt(b*x^4 + a)/x^2)/(sqrt(b) + sqrt(b*x^4 + a)/x^2))/b^(3/2) - 1/4*sqrt(b*x^4 + a)*a/((
b^2 - (b*x^4 + a)*b/x^4)*x^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^5}{\sqrt {b\,x^4+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^4)^(1/2),x)

[Out]

int(x^5/(a + b*x^4)^(1/2), x)

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sympy [A]  time = 4.17, size = 46, normalized size = 0.87 \[ \frac {\sqrt {a} x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4 b} - \frac {a \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b*x**4+a)**(1/2),x)

[Out]

sqrt(a)*x**2*sqrt(1 + b*x**4/a)/(4*b) - a*asinh(sqrt(b)*x**2/sqrt(a))/(4*b**(3/2))

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